Integration of an inverse function I need to evaluate
$$\int_0^1\frac{\sin^{-1}x}x dx.$$
I have tried integration by parts (taking $1/x$ as second function) but I am getting complex results.  Any idea on how to approach will be appreciated.  
 A: This is tricky. Do integration by parts (assuming you justify convergence at $0$). Then you're left with 
$$\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}dx.$$
Substitute $x=\sin u$ and note that you can do the definite integral
$$\int_0^{\pi/2} \ln(\sin u)\,du = \frac12\int_0^\pi \ln(\sin u)\,du$$
by the trick of substituting $u=2v$ and using the double angle formula and symmetry.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{1}{\arcsin\pars{x} \over x}\,\dd x & =
-\int_{0}^{1}{\ln\pars{x} \over \root{1 - x^{2}}}\,\dd x =
-\,{1 \over 4}\int_{0}^{1}x^{-1/2}\ln\pars{x}\pars{1 - x}^{-1/2}\,\dd x
\\[5mm] & =
\left. -\,{1 \over 4}\,\partiald{}{\nu}\int_{0}^{1}x^{\nu -1/2}
\,\pars{1 - x}^{-1/2}\,\dd x\,\right\vert_{\ \nu\ =\ 0}
\\[5mm] & =
-\,{1 \over 4}\,\partiald{}{\nu}\bracks{\Gamma\pars{\nu + 1/2}\Gamma\pars{1/2} \over \Gamma\pars{\nu + 1}}_{\ \nu\ =\ 0}
\\[5mm] & =
-\,{1 \over 4}\,\root{\pi}\,\partiald{}{\nu}\bracks{\Gamma\pars{\nu + 1/2} \over \Gamma\pars{\nu + 1}}_{\ \nu\ =\ 0}
\\[5mm] & = \bbx{{1 \over 2}\,\ln\pars{2}\,\pi} \approx 1.0888
\end{align}

Note that $\ds{\Gamma\pars{\nu + a} \sim \Gamma\pars{a} + \Gamma\pars{a}\Psi\pars{a}\nu}$ as $\ds{\nu \to 0}$ such that

\begin{align}
&{\Gamma\pars{\nu + 1/2} \over \Gamma\pars{\nu + 1}} \sim
{\Gamma\pars{1/2} \over \Gamma\pars{1}}\,
{1 + \Psi\pars{1/2}\nu \over 1 + \Psi\pars{1}\nu} \sim
\root{\pi}\bracks{1 + \Psi\pars{1 \over 2}\nu}
\bracks{1 - \Psi\pars{1}\nu}
\\[5mm] \sim &\
\root{\pi}\braces{\rule{0pt}{5mm}1 + \bracks{\Psi\pars{1 \over 2} - \Psi\pars{1}}\nu}
\quad \mbox{as}\ \nu \to 0
\\[5mm] &\
\mbox{and}\
\Psi\pars{1 \over 2} - \Psi\pars{1} =
\int_{0}^{1}{1 - t^{-1/2} \over 1 - t}\,\dd t =
2\int_{0}^{1}{t - 1 \over 1 - t^{2}}\,\dd t =
-2\int_{0}^{1}{\dd t \over 1 + t}\,\dd t = -2\ln\pars{2}
\\[5mm] &\
\mbox{such that}\
\pars{-\,{1 \over 4}\,\root{\pi}}\root{\pi}\bracks{-2\ln\pars{2}} =
\bbx{{1 \over 2}\,\ln\pars{2}\,\pi}
\end{align}
A: Just another way.
Consider
$$\sin ^{-1}(x)= \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1}$$
$$\int_0^a\frac{\sin^{-1}(x)}x\, dx=\sum_{n=0}^\infty \int_0^a \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n}}{2n+1}\,dx=a \,
   _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};a^2\right
   )$$ where appears the hypergeometric function which beautifully simplies for $a=1$ (only).
